-Two objecs ‘A’ and ‘B’ of identical mass traveling in the opposite directions w
ID: 2033097 • Letter: #
Question
-Two objecs ‘A’ and ‘B’ of identical mass traveling in the opposite directions with the speed of Va=4m/s and Vb=5m/s make a head on collision. The speed of each object after the collision if the collision is perfectly elastic isa- Va=4m/s and Vb=4m/s b- Va=4m/s and Vb=5m/s c- Va=5m/s and Vb=4m/s d- Va=5m/s and Vb=5m/s
2-A 0.24kg glider moving with a velocity of 0.6 m/s collides with and sticks to a 0.26 kg glider moving with a velocity of 0.2m/s. The loss of kinetic energy is
a-0.01 J b-0.184 J c-0.0484 J d-0.0432 J
-Two objecs ‘A’ and ‘B’ of identical mass traveling in the opposite directions with the speed of Va=4m/s and Vb=5m/s make a head on collision. The speed of each object after the collision if the collision is perfectly elastic is
a- Va=4m/s and Vb=4m/s b- Va=4m/s and Vb=5m/s c- Va=5m/s and Vb=4m/s d- Va=5m/s and Vb=5m/s
2-A 0.24kg glider moving with a velocity of 0.6 m/s collides with and sticks to a 0.26 kg glider moving with a velocity of 0.2m/s. The loss of kinetic energy is
a-0.01 J b-0.184 J c-0.0484 J d-0.0432 J
a- Va=4m/s and Vb=4m/s b- Va=4m/s and Vb=5m/s c- Va=5m/s and Vb=4m/s d- Va=5m/s and Vb=5m/s
2-A 0.24kg glider moving with a velocity of 0.6 m/s collides with and sticks to a 0.26 kg glider moving with a velocity of 0.2m/s. The loss of kinetic energy is
a-0.01 J b-0.184 J c-0.0484 J d-0.0432 J
Explanation / Answer
1.
Now it is elastic collision, so using momentum conservation
m1u1 + m2u2 = m1v1 + m2v2
u1 = 4 m/sec and u2 = -5 m/sec
m*4 + m*(-5) = m*v1 + m*v2
-m = m*v1 + m*v2
v1 + v2 = -1
another condition for elastic collision
u1 - u2 = v2 - v1
v2 - v1 = 4 - (-5)
v2 - v1 = 9
v2 = v1 + 9
Use above condition
v1 + v2 = -1
v1 + v1 + 9 = -1
v1 = -5 m/sec
v2 = -5 + 9 = 4 m/sec
Correct option is C.
Q2.
In this type of collision:
using moment conservation
Pi = Pf
m1v1 + m2v2 = M*V
where M = m1 + m2
V = (m1v1 + m2v2)/(m1 + m2)
V = (0.24*0.6 + 0.26*0.2)/(0.24 + 0.26)
V = 0.392 m/sec
Loss in KE will be
dKE = KEi - KEf
dKE = 0.5*m1*v1^2 + 0.5*m2*v2^2 - 0.5*(m1 + m2)*V^2
dKE = 0.5*0.24*0.6^2 + 0.5*0.26*0.2^2 - 0.5*(0.24 + 0.26)*0.392^2
dKE = 0.009984 = 0.01 J
Correct option is A.
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